A block of uniform cross-sectional area floats in two liquids A and B. Given densities and submerged heights, calculate displaced weights.

Here are the calculations for the floating block:

Given:
Cross-sectional area of the block, $A = 6.0 \text{ cm}^2 = 6.0 \times 10^{-4} \text{ m}^2$
Density of liquid A, $\rho_A = 800 \text{ kg/m}^3$
Density of liquid B, $\rho_B = 1000 \text{ kg/m}^3$
Height of block submerged in liquid A, $h_A = 2 \text{ cm} = 0.02 \text{ m}$
Height of block submerged in liquid B, $h_B = 2 \text{ cm} = 0.02 \text{ m}$
Total height of the block, $H_{total} = 2 \text{ cm} + 2 \text{ cm} + 2 \text{ cm} = 6 \text{ cm} = 0.06 \text{ m}$
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

i) Weight of liquid A displaced

Step 1: Calculate the volume of liquid A displaced.

$$V_A = A \times h_A$$ $$V_A = (6.0 \times 10^{-4} \text{ m}^2) \times (0.02 \text{ m})$$ $$V_A = 1.2 \times 10^{-5} \text{ m}^3$$

Step 2: Calculate the weight of liquid A displaced.

$$W_A = \rho_A V_A g$$ $$W_A = (800 \text{ kg/m}^3) \times (1.2 \times 10^{-5} \text{ m}^3) \times (9.8 \text{ m/s}^2)$$ $$W_A = 0.09408 \text{ N}$$

The weight of liquid A displaced is $\boxed{\text{0.09408 N}}$.

ii) Weight of liquid B displaced

Step 1: Calculate the volume of liquid B displaced.

$$V_B = A \times h_B$$ $$V_B = (6.0 \times 10^{-4} \text{ m}^2) \times (0.02 \text{ m})$$ $$V_B = 1.2 \times 10^{-5} \text{ m}^3$$

Step 2: Calculate the weight of liquid B displaced.

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